Question

Let \(M = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) be a \(2 \times 2\) matrix, where \(\theta = \dfrac{2\pi}{5}\), and \(I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
Which of the following options is equal to \(M^{2026}\)?

• A. \(M^2\)
• B. \(M\)
• C. \(M^{-1}\)
• D. \(I_2\)

Hint:

\(M\) is a rotation matrix by \(\theta\), so \(M^k\) is rotation by \(k\theta\), which only depends on \(k\theta \bmod 2\pi\), i.e., on \(k \bmod 5\) here. Find \(2026 \bmod 5\).

Answer 1

The Correct Option is B

Step 1: Recognize what \(M\) represents.
The matrix \(M = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) is the standard 2D rotation matrix, it rotates any vector counter-clockwise by angle \(\theta\) about the origin.

Step 2: Use the key property of rotation matrices.
Applying the rotation matrix \(k\) times in a row rotates a vector by \(k\theta\) in total, so \(M^k\) is itself the rotation matrix for angle \(k\theta\):
\[ M^k = \begin{pmatrix} \cos(k\theta) & -\sin(k\theta) \\ \sin(k\theta) & \cos(k\theta) \end{pmatrix} \]
Also, since angles that differ by a full \(2\pi\) (a complete revolution) give the exact same rotation, \(M^k\) only depends on \(k\theta\) modulo \(2\pi\).

Step 3: Compute \(2026\theta\) modulo \(2\pi\).
Here \(\theta = \dfrac{2\pi}{5}\), so:
\[ 2026\,\theta = 2026 \times \frac{2\pi}{5} = \frac{2026}{5} \times 2\pi \]
Write \(2026 = 5 \times 405 + 1\), so \(\dfrac{2026}{5} = 405 + \dfrac{1}{5}\). Then:
\[ 2026\,\theta = \left(405 + \frac{1}{5}\right) \times 2\pi = 405 \times 2\pi + \frac{2\pi}{5} \]
The term \(405 \times 2\pi\) is exactly \(405\) full revolutions, which brings us back to the same direction, so it contributes nothing extra to the rotation. What remains is just \(\dfrac{2\pi}{5} = \theta\).

Step 4: Conclude the value of \(M^{2026}\).
Since \(2026\theta\) and \(\theta\) represent the same rotation angle (modulo \(2\pi\)), we get:
\[ M^{2026} = M^{1} = M \]

Step 5: Why the other options are wrong.
(A) \(M^2\) would only be correct if \(2026 \equiv 2 \pmod 5\), but we found \(2026 \equiv 1 \pmod 5\), so this is wrong.
(C) \(M^{-1}\) is the rotation by \(-\theta\) (or equivalently \(4\theta\), since \(M^5 = I_2\) means \(M^{-1} = M^4\)), which does not match \(k \equiv 1 \pmod 5\).
(D) \(I_2 = M^0\) (or \(M^5\)) would require \(2026 \equiv 0 \pmod 5\), but \(2026\) leaves remainder \(1\), not \(0\), when divided by \(5\).

Final Answer:
\[ \boxed{M^{2026} = M} \]